• morrowcosom
In summary, you can solve this problem with an even number of transpositions if you start and end at the same permutation.
morrowcosom

## Homework Statement

(1 2 3 4 5)
(2 1 3 5 4) Write the bottom number as a product of transpositions

## The Attempt at a Solution

41352 41325 41235 42135 42315 24315 24351 21354
(2 4) (2 5) (2 3) (2 1) (3 1) (4 2) (5 1) (4 1)

This an even number of transpositions but I ended up with the number. I read somewhere that it is impossible to calculate these products with an even number of transpositions if there is an odd number of digits or vice-versa. What is up with this, or what am I doing wrong? Is there a method to finding these products, or do you just plug in random numbers after the n-1 transpositions (2, 4) (2, 5) (2, 3) (2, 1)?

I don't follow what you did. Where did 41352 come from? Are you trying to apply the transpositions below each number to get the next? If you are, you don't seem to be doing it correctly. Could you explain more clearly what the problem is and your actual work?

Also, are you saying if you have an odd number of objects, there are no permutations of those objects that is the result of an even number of transpositions? That's obviously false because you can, say, start with 5 objects and apply two transpositions, right?

Whether or not a permutation is even or odd depends on its sign; 1 if even, -1 if odd, with
$$\mbox{sgn }\pi=(-1)^{n-c(\pi)}$$
where $$c(\pi)$$ is the number of cycles in the disjoint cycle representation of $$\pi$$.

It doesn't only depend on number of digits.

So in this problem,
$$\left(\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 2 & 1 & 3 & 5 & 4 \end{array}\right) = (1,2)(4,5)(3)$$

So it's disjoint cycle rep. has 3 cycles. Therefore $$n-c(\pi)$$ will be even, so the permutation's sign will be 1.

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Whether or not a permutation is even or odd depends on its sign; 1 if even, -1 if odd, with

where is the number of cycles in the disjoint cycle representation of .

It doesn't only depend on number of digits.

So in this problem,

So it's disjoint cycle rep. has 3 cycles. Therefore will be even, so the permutation's sign will be 1.

I worked a transposition problem
(1 2 3 4 5)
(2 5 4 3 1) find the the product of the bottom sequence in transpositions.

My first step was to determine whether the number of transpositions was to be even or odd.
I presumed n=5, since this was the total numbers in the sequence. I then found disjointed cycles of (1 2 5) (3 4). I finally subtracted the number of disjointed cycles from n, to arrive at the conclusion that there should be an odd number of transpositions, since 3 is an odd number. But, I was able to solve the problem with an even number of transpositions. Is this supposed to be impossible, or is it just wrong?

25431
1) 15432 (2 1)
2) 15423 (2 3)
3) 15243 (2 4)
4) 12543 (2 5)
5) 21543 (2 1)
6) 25143 (1 5)
7) 25413 (1 4)
8) 25431 (1 3)

You're starting and ending at the same permutation, so of course you're going to find an even number of transpositions. You want to start with 12345 and find the number of transpositions needed to end at 25431.

$$\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\ 2 & 1 & 3 & 5 & 4 \end{pmatrix}$$

I see immediately that 1 and 2 are swapped, that 3 remains in the same position and that 4 and 5 are swapped. This is the same as (12)(45).

" I read somewhere that it is impossible to calculate these products with an even number of transpositions if there is an odd number of digits"
No, that is not true. All transpositions on any number of digits are either "even" (require an even number of transpositions) or "odd" (require an odd number of transpositions)- and that, for any number of digits, exactly half of the transpositons are of each kind.

I suspect that you are incorrectly remembering that, while there may be many different ways of resolving a given permutation into a product of transpositions, they are either all odd or all even.

If one way of resolving a given permutation into a product of transpostions requires an odd number of transpositions, then there cannot be a way of resolving that same permutation into an even number of transpositions.

That has nothing to do with the number of digits.

## 1. What is the difference between a permutation and a transposition?

A permutation is an arrangement of objects where the order matters, while a transposition is a type of permutation where only two objects are switched in position.

## 2. Can a permutation and a transposition be contradictory?

Yes, a permutation and a transposition can be contradictory if the transposition changes the order of objects in a way that contradicts the original permutation. This would result in a different arrangement of objects than the original permutation.

## 3. How do you identify a contradiction between a permutation and a transposition?

To identify a contradiction, you would first need to know the original permutation and the transposition. Then, you can compare the arrangement of objects in the permutation to the arrangement after the transposition has been applied. If the two arrangements are different, a contradiction exists.

## 4. What is the significance of permutations and transpositions in mathematics?

Permutations and transpositions are important concepts in combinatorics, which is the branch of mathematics that deals with counting and arranging objects. They are used to solve problems related to arrangements, combinations, and probability.

## 5. Are there any real-world applications of permutations and transpositions?

Yes, permutations and transpositions have various real-world applications, such as in cryptography, music theory, and genetics. They can also be used to solve problems related to scheduling, routing, and optimization.

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